POJ 2462 / HDU 1154 Cutting a Polygon

就这样莫名其妙的过了,不过可以确定之前都是被精度卡死了。真心受不了精度问题了。

题意:一条直线在一个不规则多边形内的长度,包括边重合部分。

首先计算出所有交点,然后按想x,y的大小进行二级排序。

然后判断相邻两点之间的线段是否在多边形内。

设两点为a,b,其重点为c。

若c在内部或边上,则线段ab必符合条件。

判断点在多边形内的方法可以参考下面的链接:

http://www.cnblogs.com/hhyypp/archive/2011/12/05/2276984.html

#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <queue>
#include <cmath>
#include <algorithm>

#define LL long long
#define EPS (1e-8)

const double Double_PI = 2.0*acos(-1.0);

using namespace std;

struct P
{
    double x,y;
}sp[100100],p[100100];

int top;

bool cmp(P p1,P p2)
{
    if(fabs(p1.x - p2.x) < EPS)
    {
        return p1.y < p2.y;
    }
    return p1.x < p2.x;
}

double X_Mul(P a1,P a2,P b1,P b2)
{
    P v1 = {a2.x-a1.x , a2.y-a1.y},v2 = {b2.x-b1.x , b2.y-b1.y};

    return (v1.x*v2.y-v1.y*v2.x);
}

P Line_Cross_Position(P l1,P l2,P s1,P s2)
{
    P cp = s1;

    double t = fabs(X_Mul(l1,l2,l1,s1)) / fabs(X_Mul(l1,l2,s2,s1));

    cp.x += t*(s2.x-s1.x);
    cp.y += t*(s2.y-s1.y);

    return cp;
}


double Cal_Angle(P a1,P a2,P b1,P b2)
{
    P v1 = {a2.x-a1.x,a2.y-a1.y},v2 = {b2.x-b1.x,b2.y-b1.y};

    if(fabs(X_Mul(a1,a2,b1,b2)) < EPS)
        return 0;

    if(X_Mul(a1,a2,b1,b2) >= EPS)
        return acos((v1.x*v2.x+v1.y*v2.y)/(sqrt( (v1.x*v1.x + v1.y*v1.y)*(v2.x*v2.x + v2.y*v2.y) )));
    return -acos((v1.x*v2.x+v1.y*v2.y)/(sqrt( (v1.x*v1.x + v1.y*v1.y)*(v2.x*v2.x + v2.y*v2.y) ) ) );
}

double Cal_Point_Dis(P p1,P p2)
{
    return sqrt( (p1.x-p2.x)*(p1.x-p2.x) + (p1.y-p2.y)*(p1.y-p2.y) );
}

double Point_In_Segment(P p,P p1,P p2)
{
    if( fabs(Cal_Point_Dis(p1,p2) - Cal_Point_Dis(p,p1) - Cal_Point_Dis(p,p2) ) < EPS)
        return true;
    return false;
}

bool Is_Point_In_Polygon(P tp,int n)
{
    int i;

    double asum = 0;

    for(i = 0;i < n; ++i)
    {
        asum += Cal_Angle(tp,p[i],tp,p[i+1]);
    }

    if(fabs(fabs(asum) - Double_PI) < EPS)
        return true;
    for(i = 0;i < n; ++i)
    {
        if(Point_In_Segment(tp,p[i],p[i+1]) )
            return true;
    }
    return false;
}

double Length_In_Polygon(P pa,P pb,int n)
{
    double lsum,xm1,xm2;
    int i;
    P temp;

    for(i = 0;i < n; ++i)
    {
        xm1 = X_Mul(pa,pb,pa,p[i]);
        xm2 = X_Mul(pa,pb,pa,p[i+1]);

        if(xm1*xm2 < -EPS)
        {
            sp[top++] = Line_Cross_Position(pa,pb,p[i],p[i+1]);
        }
        else
        {
            if(fabs(xm1) < EPS)
                sp[top++] = p[i];
            if(fabs(xm2) < EPS)
                sp[top++] = p[i+1];
        }
    }

    sort(sp,sp+top,cmp);

    for(i = 1,lsum = 0;i < top; ++i)
    {
        temp.x = (sp[i].x + sp[i-1].x)/2;
        temp.y = (sp[i].y + sp[i-1].y)/2;

        if(Is_Point_In_Polygon(temp,n))
        {
            lsum += Cal_Point_Dis(sp[i],sp[i-1]);
        }
    }
    return lsum;
}

int main()
{
    int i,n,m;
    P pa,pb;

    while( scanf("%d %d",&n,&m) && (n||m) )
    {
        for(i = 0;i < n; ++i)
        {
            scanf("%lf %lf",&p[i].x,&p[i].y);
        }

        p[n] = p[0];

        while(m--)
        {
            scanf("%lf %lf %lf %lf",&pa.x,&pa.y,&pb.x,&pb.y);

            top = 0;

            printf("%.3lf
",Length_In_Polygon(pa,pb,n));
        }
    }
    return 0;
}
原文地址:https://www.cnblogs.com/riskyer/p/3400345.html